Question: Simplify and expand the following expression: $ \dfrac{2}{y - 4}- \dfrac{5}{4y - 4}+ \dfrac{2y}{y^2 - 5y + 4} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the second term: $ \dfrac{5}{4y - 4} = \dfrac{5}{4(y - 1)}$ We can factor the quadratic in the third term: $ \dfrac{2y}{y^2 - 5y + 4} = \dfrac{2y}{(y - 4)(y - 1)}$ Now we have: $ \dfrac{2}{y - 4}- \dfrac{5}{4(y - 1)}+ \dfrac{2y}{(y - 4)(y - 1)} $ The least common multiple of the denominators is: $ (y - 4)(y - 1)$ In order to get the first term over $(y - 4)(y - 1)$ , multiply by $\dfrac{4(y - 1)}{4(y - 1)}$ $ \dfrac{2}{y - 4} \times \dfrac{4(y - 1)}{4(y - 1)} = \dfrac{8(y - 1)}{(y - 4)(y - 1)} $ In order to get the second term over $(y - 4)(y - 1)$ , multiply by $\dfrac{y - 4}{y - 4}$ $ \dfrac{5}{4(y - 1)} \times \dfrac{y - 4}{y - 4} = \dfrac{5(y - 4)}{(y - 4)(y - 1)} $ In order to get the third term over $(y - 4)(y - 1)$ , multiply by $\dfrac{4}{4}$ $ \dfrac{2y}{(y - 4)(y - 1)} \times \dfrac{4}{4} = \dfrac{8y}{(y - 4)(y - 1)} $ Now we have: $ \dfrac{8(y - 1)}{(y - 4)(y - 1)} - \dfrac{5(y - 4)}{(y - 4)(y - 1)} + \dfrac{8y}{(y - 4)(y - 1)} $ $ = \dfrac{ 8(y - 1) - 5(y - 4) + 8y} {(y - 4)(y - 1)} $ Expand: $ = \dfrac{8y - 8 - 5y + 20 + 8y}{4y^2 - 20y + 16} $ $ = \dfrac{11y + 12}{4y^2 - 20y + 16}$